The moment of inertia of a ring about an axis passing through the centre and perpendicular to its plane is l. It is rotating with angular velocity $\omega$. Another identical ring is gently placed on it, so that their centres coincide. If  both the rings are rotating  about the same axis, then loss in kinetic energy is 

A) $\frac{l\omega^{2}}{2}$

B) $\frac{l\omega^{2}}{4}$

C) $\frac{l\omega^{2}}{6}$

D) $\frac{l\omega^{2}}{8}$


Option B


  According to the law of conservation of angular momentum, 

    l $\omega$  = constant

 Now, according  to the question,

$l_{1}\omega_{1}=l_{2}\omega_{2}$ or  $ k\omega= (2l) \omega_{2}$


 New kinetic energy =  $[\frac{1}{2}l_{2}\omega_{2}^{2}]= \frac{1}{2}(2l)\times (\frac{\omega}{2})^{2}= \frac{l\omega^{2}}{4}$

 loss in kinetic energy (KL)= Ki  - Kj

 =  $\frac{1}{2}l_{}\omega_{}^{2}-  \frac{l\omega^{2}}{4}=\frac{l\omega^{2}}{4}$